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Chapter 6: Problem 103

When \(1.034 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) are burned in a constant-volume bomb calorimeter at \(298 \mathrm{~K}, 41.56 \mathrm{~kJ}\) of heat are evolved. Calculate \(\Delta E\) and \(\Delta H\) for the reaction on a molar basis.

Short Answer

Expert verified
The change in internal energy (ΔE) and the change in enthalpy (ΔH) of the combustion of naphthalene on a molar basis are both -5155 kJ/mol.

Step by step solution

01

Conversion of gram to mole

First, convert the mass of naphthalene used for combustion to mole using its molar mass (128.17 g/mol).\[\text{{Number of moles}} = \frac{{1.034 g}}{{128.17 g/mol}} = 0.00806 mol\]
02

Calculation of ΔE

Since ΔE=q (where q is the heat evolved), divide the total heat evolved by the number of moles to get ΔE on a per mole basis.\[\Delta E = \frac{{41.56 kJ}}{{0.00806 mol}} = -5155 kJ/mol.\]Remember that q is negative since heat is evolved (exothermic process). So, the value of ΔE will also be negative.
03

Conclusion on ΔH

For a constant volume process, such as this one in a bomb calorimeter, the change in volume (ΔV) is zero. Because of this, ΔH is essentially the same as ΔE. Hence,\[\Delta H = \Delta E = -5155 kJ/mol.\]

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Most popular questions from this chapter

Consider this reaction: $$ \begin{aligned} 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow & 4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \\ \Delta H &=-1452.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ What is the value of \(\Delta H\) if (a) the equation is multiplied throughout by \(2,\) (b) the direction of the reaction is reversed so that the products become the reactants and vice versa, (c) water vapor instead of liquid water is formed as the product?

I he enthaipy of combustion benzo1c acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be \(-3226.7 \mathrm{~kJ} / \mathrm{mol}\). When \(1.9862 \mathrm{~g}\) of benzoic acid are burned in a calorimeter, the temperature rises from \(21.84^{\circ} \mathrm{C}\) to \(25.67^{\circ} \mathrm{C}\). What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly \(2000 \mathrm{~g} .)\)

You are given the following data: $$ \begin{array}{c} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \quad \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \quad \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \\ \Delta H^{\circ}=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}(g)+\operatorname{Br}(g) \longrightarrow \operatorname{HBr}(g) $$

Calculate the work done in joules when 1.0 mole of water vaporizes at 1.0 atm and \(100^{\circ} \mathrm{C}\). Assume that the volume of liquid water is negligible compared with that of steam at \(100^{\circ} \mathrm{C}\), and ideal gas behavior.

Consider the following data: $$ \begin{array}{lcc} \text { Metal } & \text { Al } & \text { Cu } \\ \hline \text { Mass (g) } & 10 & 30 \\ \text { Specific heat }\left(\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right) & 0.900 & 0.385 \\ \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & 40 & 60 \end{array} $$ When these two metals are placed in contact, which of the following will take place? (a) Heat will flow from Al to Cu because Al has a larger specific heat. (b) Heat will flow from Cu to Al because Cu has a larger mass. (c) Heat will flow from Cu to Al because Cu has a larger heat capacity. (d) Heat will flow from Cu to Al because Cu is at a higher temperature. (e) No heat will flow in either direction.
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